Show how would you join three resistors, each of resistance 9 $\Omega$ so that the equivalent resistance of the combination is (i) 13.5 $\Omega$, (ii) 6 $\Omega$?

To solve this problem, we need to use the concept of series and parallel combinations of resistors. (i) To get an equivalent resistance of 13.5 Ω, we need to connect two resistors in parallel and then connect the third resistor in series with the parallel combination. Let's calculate the equivalent resistance of the parallel combination: 1/Rp = 1/R1 + 1/R2 = 1/9 + 1/9 = 2/9 Rp = 9/2 = 4.5 Ω Now, we connect the third resistor in series with the parallel combination: Rs = R3 + Rp = 9 + 4.5 = 13.5 Ω (ii) To get an equivalent resistance of 6 Ω, we need to connect all three resistors in parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 = 1/9 + 1/9 + 1/9 = 3/9 Rp = 9/3 = 3 Ω However, this is not equal to 6 Ω. To get 6 Ω, we need to connect two resistors in series and then connect the third resistor in parallel with the series combination: Rs = R1 + R2 = 9 + 9 = 18 Ω Now, we connect the third resistor in parallel with the series combination: 1/Rp = 1/Rs + 1/R3 = 1/18 + 1/9 = 3/18 Rp = 18/3 = 6 Ω